how to open file dialog in c# code example
Example 1: c# open file dialog
OpenFileDialog dialog = new OpenFileDialog();
if (DialogResult.OK == dialog.ShowDialog())
{
string path = dialog.FileName;
}
Example 2: open file dialog c#
var fileContent = string.Empty;
var filePath = string.Empty;
using (OpenFileDialog openFileDialog = new OpenFileDialog())
{
openFileDialog.InitialDirectory = "c:\\";
openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
openFileDialog.FilterIndex = 2;
openFileDialog.RestoreDirectory = true;
if (openFileDialog.ShowDialog() == DialogResult.OK)
{
filePath = openFileDialog.FileName;
var fileStream = openFileDialog.OpenFile();
using (StreamReader reader = new StreamReader(fileStream))
{
fileContent = reader.ReadToEnd();
}
}
}
MessageBox.Show(fileContent, "File Content at path: " + filePath, MessageBoxButtons.OK);